Math Component:

A cannonball is shot upward from the upper deck of a fort with an initial velocity of 192 feet per second.  The deck is 32 feet above the ground. 

Quadratic Model: h = -16t²+192t+32

How high does the cannonball go?  608 feet
How long is the cannonball in the air? 12.16 seconds
Steps for Solving:
1. Plug 192 feet per second in for v₀ and 32 feet for h₀
2. Plug formula into calculator and graph
3. Go to the table and find the vertex of the graph (this is how high the cannonball went)
4. Using the quadratic formula, find the positive x-intercept (this is how long the cannonball was in the air)
5. Check your work


Launch Angle Choice:


We chose to use a 50 degree angle because we used predicted numbers for our cannon using a formula Mrs. Tyrpak provided for us (it was an altered version of the quadratic formula). We believe that if design permits the 50 degree angle will yield the most fruitful results.


Gas Law:


The gas law we used was Gay-Lussacs. The components in this cannon project are pressure and temperature. We aren't exactly sure about the exact numbers to use because we are unsure of the environmental conditions on the day we will be launching.


Design Choice:


We chose the design we did for multiple reasons. We chose to make a two chamber barrel because we will believe that it will increase the pressure exerted. We used a sturdier base so that wind conditions won't alter the cannon angle. We also made sure to place the weight in a way the cannon won't tip over.